We are asked to solve the equation $(x-2)(x+1) = \frac{2x+19}{2}$ for $x$ using the quadratic formula.

1. Problem Description

We are asked to solve the equation $(x-2)(x+1) = \frac{2x+19}{2}$ for $x$ using the quadratic formula.

2. Solution Steps

First, expand the left side of the equation:

$$(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$$

So the equation is now:

$$x^2 - x - 2 = \frac{2x+19}{2}$$

Multiply both sides by 2 to eliminate the fraction:

$$2(x^2 - x - 2) = 2x + 19$$

$$2x^2 - 2x - 4 = 2x + 19$$

Move all terms to the left side to obtain a quadratic equation in standard form:

$$2x^2 - 2x - 4 - 2x - 19 = 0$$

$$2x^2 - 4x - 23 = 0$$

Now, we can use the quadratic formula to solve for $x$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 2$, $b = -4$, and $c = -23$.

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-23)}}{2(2)}$$

$$x = \frac{4 \pm \sqrt{16 + 184}}{4}$$

$$x = \frac{4 \pm \sqrt{200}}{4}$$

Since $200 = 100 \cdot 2$, we have $\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$.

$$x = \frac{4 \pm 10\sqrt{2}}{4}$$

Divide both terms in the numerator by 2:

$$x = \frac{2 \pm 5\sqrt{2}}{2}$$

So the two solutions are:

$$x = \frac{2 + 5\sqrt{2}}{2} \quad \text{and} \quad x = \frac{2 - 5\sqrt{2}}{2}$$

3. Final Answer

$\frac{2+5\sqrt{2}}{2}, \frac{2-5\sqrt{2}}{2}$